YodaEmbedding
commented
4 days ago For each $i$, there is a discrete probability distribution $p_i$, i.e.,
$$\int_{\mathbb{R}} p_i(t) \, dt = 1.$$
There are many such probability distributions -- one for each element in $\hat{y}$.
Each element $\hat{y}_i$ is encoded using its corresponding $p_i$. We can measure the "likelihood" $l_i$ for each element:
$$l_i = p_i(\hat{y}_i)$$
...and since the rate is the negative log-likelihood, the bit cost of the $i$-th element is:
$$R_i = -\log_2 l_i$$
The total rate cost is then:
$$R = \sum_i R_i$$
...which can be averaged over $N$ pixels to obtain the bpp.
For a single fixed encoding distribution $p$, the average rate cost for encoding a single symbol that is drawn from the same distribution $p$ is:
$$R = \sum_t - p(t) \, \log p(t)$$
But this is not what we're doing. What we're actually interested in is the cross-entropy. That is the average rate cost for encoding a single symbol drawn from the true distribution $\hat{p}$:
$$R = \sum_t - \hat{p}(t) \, \log p(t)$$
To be consistent with our notation above, we should also sprinkle in some $i$ s:
$$R_i = \sum_t - \hat{p}_i(t) \, \log p_i(t)$$
In our case, we know exactly what $\hat{p}$ is...
$$\hat{p}_i(t) = \delta[t - \hat{y}_i] = \begin{cases} 1 & \text{if } t = \hat{y}_i \ 0 & \text{otherwise} \end{cases} $$
If we plug this into the earlier equation, the rate cost for encoding the $i$-th element becomes:
$$R_i = -\log p_i(\hat{y}_i)$$
Hi, it's an excellent work. I have recently been working in the area of learned compression. I have already checked issue #306, but I am still confused about the actual meaning of
y_likelihoods:The shape of
y_likelihoodsis equal to the shape ofy, and I thought each element ofy_likelihoodsrepresents the probability $p(y_i)$, where $p(y_i)$ is the probability that the symbol $y_i$ appears. However, when I run the code below from the demo,it produces an output where the elements of
y_likelihoodsare close to 1 everywhere. Isn't this contradictory? Shouldn't the sum of y_likelihoods is 1? What is the actual meaning ofy_likelihoods?The calculation for bpp is $\sum -log_2(p(y_i))/N$, but the information entropy is actually $\sum - p(y_i) log_2(p(y_i))$, I just wandering where did the $p(y_i)$ go? Is it because $\sum -log_2(p(y_i))/N$ is the upper bound of $\sum - p(y_i)\log_2(p(y_i))$ from the inequality below?